Talk:Kinetic theory of gases

Untitled

This article contains numerous errors along with non-standard terminology. Ideal gasses can be both relativistic and quantum. I think the author intends to restrict his derivation to non-relativistic classical ideal gasses. Even in this case, most of the assumptions, such as equal mass particles, are not required. The "noninteracting" assumption does not imply either 1 or 2 of the points listed under it, and point 3 depends only on one of the above. Someone needs to rewrite this. 2607:F388:101C:0:7944:8FAD:2F67:6CFC (talk) 03:40, 21 January 2018 (UTC)[reply]

Postulates

These molecules my toe are in constant, random motion.

A particle can achive a velocity of 0 m/s, so i would deny it is in constant motion I suggest: The molecules move in random and chaotic motion. The velocities that the molecules can attain, cover the complete range between 0 and infinity and are distributed according the Maxwell-Bolzmann frequency distribution function.

Huisman 15:06, 2006 May 8 (GMT)

consider water as an example and how it changes for the states gas, liquid and solid. Think about all the macroscopic and microscopic things that happen and this theory will begin to make since. —Preceding unsigned comment added by 138.88.75.90 (talk) 23:00, 31 January 2008 (UTC)[reply]

Not entirely true, all molecules in a gas always have movement, however minuscule, above the temperature of -273.15°C. 10/15/09 3:30 EST —Preceding unsigned comment added by 199.76.171.61 (talk) 19:30, 15 October 2009 (UTC)[reply]

LIQUIDS AND SOLIDS

Why do you seem to think that kinetic energy only applies to gasses? —Preceding unsigned comment added by 86.161.65.69 (talk) 18:16, 1 June 2009 (UTC)[reply]

Kinetic theory (of gases)

We have an article called Kinetic theory of gases that redirects to an article called Kinetic theory. But the Kinetic theory article only seems to refer to gases! Shouldn't we either put the content under Kinetic theory of gases or add information about solids and liquids?

Brianjd 05:55, 2004 Nov 13 (UTC)

Agreed; there is also a kinetic theory of solids article. David Hollman (Talk) 15:56, 22 September 2010 (UTC)[reply]

Is this article really helpful to anyone at all????

All I see is a bunch of formulas, separated by the word "thus".

Its like this, people who can understand this (not to mention would even be slightly interested) dont need to read it. They already know it. But people who want to learn, would not find their inspiration here.

Too much equations... save it for your teachers, not for the audience. — Preceding unsigned comment added by 108.39.122.43 (talk) 22:59, 29 October 2012 (UTC)[reply]

Magnetohydrodynamics

There is no discussion of the Kinetic theory for MHD (magnetohydrodynamics) which is one of the largest branches of Kinetic theory and used in Plasma Physics.

Pressure

The kinetic molecular theory has to do with the movement and collision of molecules. Molecules have to collide for a reaction to happen. The energy of the collision breaks the old chemical bonds so a reaction can happen and new bonds can form. So it's not only about gas particles.

The discussion of pressure is beyond me at the moment but I think it should be expanded to deal with any shaped container, not just a cube. Brianjd 05:39, 2004 Nov 11 (UTC)

The derivation shown is a widely accepted standard derivation. To use a cube is traditional and (as far as I know) trivial. sconzey 14:08, 2005 Sept 16 (GMT)

The notion TotalForce as it is used in the derivation is not correct or at least not consistent with the physical notion "force" which is a vector. In this sense the total force exerted in x direction is zero! The correct derivation is that if there is no preferred direction the averages of v_x^2, v_y^2 and v_z^2 all must be equal and so also equal to the average of 1/3*v^2. Cede69 14:49, 24 August 2006 (UTC)[reply]

Yes and no. You are right that the container as a whole experiences no net force; it remains stationary. I added a bit of clarification to acknowledge this point. However, the forces the derivation considers are all along "outward" vectors normal to the surface of the container. This is a perfectly sensible notion of force, and is perfectly compatible with the notion of a force as a vector. For a cube, it is easy to add up the outward forces on each of the six sides, but in more advanced derivations, the total force experienced by more complex surfaces can be calculated using calculus. Consideration of force as a sum of an infinitely large set of vectors is typical of college-level treatment of this type of theory. (Or at least it is at MIT.) -- Beland 06:58, 7 April 2007 (UTC)[reply]

"Pressure is explained by kinetic theory as arising from the force exerted by liquid molecules impacting on the walls of the container, which shows that the molecules of liquid would need less energy at the surface of the liquid to leave. Consider a gas of N molecules...."suddenly..from gasses to liquid..i coudn't get it.. —Preceding unsigned comment added by 59.94.104.6 (talk) 17:05, 11 December 2010 (UTC)[reply]

There is a conceptual error in Pressure deduction. Newtons ecuation F=dp/dt is used, and dt estimated from mean bouncing-back frequency of a molecule, but that is not the meaning of the dt. The right deduction must come from J, the integral of force in time, where a mean force is estimated based bouncing-back frequency times N. See the French version. The mean idea is that each bouncing is randomly distributed on time, so de sum of all of them gives you a constant mean force equal to J/Dt times N. Guilloip (talk) 20:36, 8 May 2017 (UTC)[reply]

Flucluation dissipation theorem

The article should link to the fluctuation dissipation theorem. (AC)

Assumptions

Do the equations not assume that the particles do *not* collide with eachother? sconzey 14:08, 2005 Sept 16 (GMT)

I think they may assume elastic collisions. -- Beland 06:43, 7 April 2007 (UTC)[reply]

The current discussion does not include collisions but the kinetic theory of gases also can be modified to include collisions between molecules and the resulting formulas are the same. Same is true for assuming that the container is cubic, i.e., non-cubic containers can also be shown to yield the same results. But those derivations are difficult and complex and not very encyclopedic. However the inclusion of attractive forces between molecules or the inclusion of vibrational and rotational degrees of freedom does change things. KeeYou Flib (talk) 17:35, 9 June 2021 (UTC)[reply]

Improvements

The discussion should include calculations of viscosity and heat conduction for ideal gases and explain how the kinetic theory is rigorously derived from the Boltzmann equation. Mention should also be made of the Chapman-Enskog approach and Grad's moment approach which apply kinetic theory to non-ideal gasses Jim McElwaine 09:52, 28 April 2006 (UTC)[reply]

Proposal to replace the symbol "N" by "n" to be in line with other articles like Ideal_gas_law —Preceding unsigned comment added by Wiki-joost (talk • contribs) 13:12, 10 April 2009 (UTC)[reply]

   It is not the same. N = number of molecules. n=N/Na= number of mols (An Avogadro Number of Molecules).  You must multiply and divide N by Na to get the ideal gases equation written with n.  — Preceding unsigned comment added by Guilloip (talk • contribs) 20:42, 8 May 2017 (UTC)[reply]

Moving molecules - why?

These molecules are in constant, random motion. The rapidly moving particles constantly collide with each other and with the walls of the container. - why are they moving? --Palnatoke 21:56, 9 February 2007 (UTC)[reply]

The simple answer is that there is nothing to stop them from moving; they experience largely elastic collisions. What got them moving in the first place is a question about the nature of the Big Bang which I'm not sure science has answered yet. -- Beland 06:43, 7 April 2007 (UTC)[reply]

I'd say that the gas is implicitly assumed to be in a state of thermal equilibrium. What that means is that the distribution of velocities is finite because the temperature of the system is finite. And the temperature is finite because enough energy (First Law of Thermodynamics ) entered the system at some point in its history to bring it to a single, well-defined equilibrium state in a manner dictated by the Second Law of Thermodynamics and the Third Law of Thermodynamics. KeeYou Flib (talk) 04:24, 19 June 2021 (UTC)[reply]

Illustration requests

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It would be very helpful to have one or more illustrations using the traditional idealized hard spheres or points. -- Beland 06:46, 7 April 2007 (UTC)[reply]

The GIF image almost crashed my Safari browser. I have a good (10mbps) internet connection, so I presume that the image has been poorly coded. Is it possible for someone to simplify this to use less bandwidth? --Info 21:56, 25 January 2010 (UTC)[reply]

Comments

This is a very well made article. This is truly excellent. —Preceding unsigned comment added by 68.226.25.53 (talk) 20:52, 23 May 2009 (UTC)[reply]

This is a Classical Mechanics Interpretation

I'm curious to know if pressure can be defined as 2/3 the kinetic energy when the velocity of the particles approaches the speed of light, and confined into regions that are of quantum length scales. —Preceding unsigned comment added by 68.32.160.9 (talk) 04:25, 10 October 2010 (UTC)[reply]

Of course not. This is a nonrelativistic model. KeeYou Flib (talk) 18:12, 9 June 2021 (UTC)[reply]

Particles with the same mass?

In the postulates there is a statement "These particles have the same mass". Surely this is not correct, a major value of kinetic theory was the discovery that the particle energy was independent of its mass. --Damorbel (talk) 21:29, 8 December 2011 (UTC)[reply]

You're right to point that out. In the derivation presented here the masses are assumed to be the same but this can easily be generalized. I'll see if I can correct that in the postulates. KeeYou Flib (talk) 18:14, 9 June 2021 (UTC)[reply]

Okay, done! KeeYou Flib (talk) 18:42, 9 June 2021 (UTC)[reply]

challenge

"Thus, the product of pressure and volume per mole is proportional to the average (translational) molecular kinetic energy."

PV = 2/3 K

The equation makes no reference to 'per mole', and Nk_B cancels out during the derivation, so I think this is valid without that qualification. Wrong or right?

173.25.54.191 (talk) 04:52, 26 June 2013 (UTC)[reply]

This article, as almost all articles connected with thermodynamics, is truly lamentable; why does it only deal with monatomic ('ideal gas') particles? This excludes enegy in particle motions e.g. vibration and rotation ('degrees of freedom). Such an exclusion completely vitiates the usefulness of the article. --Damorbel (talk) 07:27, 26 June 2013 (UTC)[reply]

I also don't like how the meaning of $m$ seems to change from "Collisions with container" to "Speed of molecules" with no further notice...130.83.66.46 (talk) 10:05, 20 February 2017 (UTC)[reply]

This seems to have been fixed, kudos to whoever took care of it. KeeYou Flib (talk) 18:44, 9 June 2021 (UTC)[reply]

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Physical Science Study 1

Kinetic energy in gas particles is beautiful Michael Mazhambe (talk) 15:36, 20 February 2017 (UTC)[reply]

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Garbled edit summary

In summarising this edit my finger inadvertently strayed onto the return key, thereby saving the edit. (A hazard of editing on a phone.) The summary should read: Unlink particles article: not helpful as it introduces more potential meanings rather than clarifying what's meant here. Musiconeologist (talk) 22:59, 11 March 2024 (UTC)[reply]

Source of the factor 3 - area or velocity?

The article currently accounts for the factor "3" by assuming equal areas in all directions (A_x, A_y, A_z). I was taught that the reason is that v_t² = v_x² + v_y² + v_z². This still makes more sense to me. The gas laws do not depend on the shape of the container. The equipartition of energies means that the velocities are uniformly distributed in space. 2003:E4:2F46:9300:5968:E6D3:B394:D72B (talk) 14:19, 5 April 2024 (UTC)[reply]